Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
struct city {
  int peo[10];
} A[100100], D[100100][17];
int tim;
int P[100100][17], in[100100], out[100100];
vector<int> g[100100];
city zero() {
  city tmp;
  memset(tmp.peo, 0, sizeof(tmp.peo));
  return tmp;
}
city merge(city a, city b) {
  city tmp = zero();
  int l = 0, r = 0, i;
  for (i = 0; i < 10; i++) {
    if (a.peo[l] == b.peo[r]) {
      tmp.peo[i] = a.peo[l];
      l++;
      r++;
    } else if (a.peo[l] == 0 || (b.peo[r] != 0 && b.peo[r] < a.peo[l])) {
      tmp.peo[i] = b.peo[r];
      r++;
    } else {
      tmp.peo[i] = a.peo[l];
      l++;
    }
  }
  return tmp;
}
void dfs(int p, int u) {
  tim++;
  in[u] = tim;
  for (int i = 1; i <= 16; i++) {
    P[u][i] = P[P[u][i - 1]][i - 1];
    if (P[u][i]) D[u][i] = merge(D[u][i - 1], D[P[u][i - 1]][i - 1]);
  }
  for (int i = 0; i < g[u].size(); i++) {
    int to = g[u][i];
    if (to == p) continue;
    D[to][0] = A[u];
    P[to][0] = u;
    dfs(u, to);
  }
  tim++;
  out[u] = tim;
}
void add(int u, int ind) {
  if (A[u].peo[9] == 0 || A[u].peo[9] > ind) A[u].peo[9] = ind;
  for (int i = 9; i >= 1; i--) {
    if (A[u].peo[i - 1] == 0 || A[u].peo[i - 1] > A[u].peo[i])
      swap(A[u].peo[i - 1], A[u].peo[i]);
  }
}
city LCA(int u, int v) {
  city tmp = merge(A[u], A[v]);
  for (int i = 16; i >= 0; i--) {
    if (P[u][i]) {
      if (in[P[u][i]] > in[v] || out[v] > out[P[u][i]]) {
        tmp = merge(tmp, D[u][i]);
        u = P[u][i];
      }
    }
    if (P[v][i]) {
      if (in[P[v][i]] > in[u] || out[u] > out[P[v][i]]) {
        tmp = merge(tmp, D[v][i]);
        v = P[v][i];
      }
    }
  }
  if (in[u] > in[v] || out[v] > out[u]) {
    if (in[v] > in[u] || out[u] > out[v]) tmp = merge(tmp, D[v][0]);
  }
  return tmp;
}
int main() {
  int n, m, q, a, b, v, u, i, ans;
  while (cin >> n >> m >> q) {
    tim = 0;
    for (i = 1; i <= n; i++) g[i].clear();
    for (i = 1; i < n; i++) {
      scanf("%d%d", &a, &b);
      g[a].push_back(b);
      g[b].push_back(a);
    }
    for (i = 1; i <= n; i++) A[i] = zero();
    for (i = 1; i <= m; i++) {
      scanf("%d", &a);
      add(a, i);
    }
    dfs(0, 1);
    while (q--) {
      scanf("%d%d%d", &u, &v, &a);
      city tmp = LCA(v, u);
      ans = 0;
      for (i = 0; i < 10 && ans < a; i++) {
        if (tmp.peo[i]) ans++;
      }
      printf("%d", ans);
      for (i = 0; i < 10 && ans; i++) {
        if (tmp.peo[i]) {
          printf(" %d", tmp.peo[i]);
          ans--;
        }
      }
      printf("\n");
    }
  }
}